Sunday, June 29, 2014

Parametric curves and second derivatives: methods

Let us know come to face the problem of computing the second derivative for functions defined through a pair of parametric equations:
Utilizing the results of the previous theorem on the first derivative, we can proceed -after having obtained the first derivative $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ as a function of the parameter $t$- to the computation of the second derivative either as
\begin{equation} \label{secder1st}
\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt}
\end{equation}
where $y'=\frac{dy}{dx}$, or as
\begin{equation} \label{secder2nd}
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dt}\Big[ \frac{dy}{dx} \Big] \cdot \frac{dt}{dx}
\end{equation}Remarks:1. Notice that -according to a previous remark in the computation of the first derivative- if we wish to apply \eqref{secder2nd}, a suitable partition of the domain $E \subseteq \mathbb{R}$ of $x=f(t)$, must be considered in order for $x=f(t)$ to be bijective ("1-1"). Thus, we will have $t=f_{i}^{-1}(x)$ (in the corresponding interval in $E$'s partition) and the correct formula for $\frac{dt}{dx}$ must be replaced.2. Notice that in general $$\frac{d^{2}y}{dx^{2}} \neq \frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$$

It is clear that $x=sint$ is invertible in the domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus we can directly apply \eqref{secder2nd}:

Since $\frac{dx}{dt}=cost$ and $\frac{dy}{dt}=-2sin2t$, we have:
$$
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2sin2t}{cost} = \frac{-4 sint cost}{cost} = -4 sint
$$
therefore:
$$
\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\Big[ \frac{dy}{dx} \Big] = \frac{d}{dx}(-4 sint) = \frac{d}{dt}(-4 sint) \cdot \frac{dt}{dx} = (-4 cost) \cdot (\frac{1}{cost}) = -4
$$
where we have made use of the fact that $x=sint \Leftrightarrow t=arcsinx$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and thus $\frac{dt}{dx} = (arcsinx)' = \frac{1}{cost} = \frac{1}{\sqrt{1-x^{2}}}$ for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$. (see previous post on the derivatives of the inverse trigonometric functions).
Notice that $\frac{dx}{dt} \Big|_{\pm \frac{\pi}{2}} = \frac{dy}{dt} \Big|_{\pm \frac{\pi}{2}}=0$ so these points are singular points and the above result does not apply at these points. Can you figure out what is happening at these points ? (plotting a graph of the parametric equations will probably help you understand the behavior at these singular points).

This curve (and the functions defined by it) can be equivalently described in implicit form by the equation $y^{2}=x^{3}$ (square $y(t)$ and cube $x(t)$ to see this!).
The graph of these parametric equations consists of two branches: the upper branch corresponding to the function $y=x^{3/2}$ while the lower branch is the graph of the function $y=-x^{3/2}$. These two branches meet at the origin, which corresponds to the value $t=0$.
Since $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^{2}$ we can clearly see that the origin corresponds to a singular point of the graph.
We can readily apply the theorem to compute the first derivative (for either branch):
$$
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{d}{dt}(t^{3})}{\frac{d}{dt}(t^{2})} = \frac{3t^{2}}{2t} = \frac{3}{2}t
$$
while we will follow \eqref{secder1st} to compute the second derivative:
$$
\frac{d^{2}y}{dx^{2}} = \frac{dy'}{dx} = \frac{dy'/dt}{dx/dt} = \frac{\frac{d}{dt}(3t/2)}{\frac{d}{dt}(t^{2})} = \frac{3}{4t}
$$
Can you apply \eqref{secder2nd}, after suitably dividing the domain $(-\infty, \infty)$, to obtain the same results?
Can you figure out what is happening at the singular point $O(0,0)$ ?